# CAT 2017 QA Slot 1 | Previous Year Questions

**1. CAT 2017 QA Slot 1 | Arithmetic - Percentage**

Arun's present age in years is 40% of Barun's. In another few years, Arun's age will be half of Barun's. By what percentage will Barun's age increase during this period?

Answer: 20

**Explanation** :

Let Barun’s age be 10x. Arun’s age is 4x.

The difference of these ages in 6x, a constant.

When Arun’s age is 50% of Barun’s age, this difference also would be 50% i.e., Barun’s age, at that stage would be 12x.

It would be increase by 20%.

**Alternately**,

Let Barun’s age be 10x. Arun’s age is 4x.

In t years, Arun's age will be 50% or half of Barun's age

⇒ (4x + t) = 1/2 × (10x + t)

⇒ t = 2x

∴ Barun's age will become 12x.

⇒ Required % = (12x - 10x)/10x × 100% = 20%

Hence, 20.

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**2. CAT 2017 QA Slot 1 | Arithmetic - Time & Work**

A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

Answer: 15

**Explanation** :

Let the number of days required to complete the job be n.

1 person works on day 1, 2 on day 2, 3 on day 3, …. n on day n.

**Using Unitary Method:**

Each person has the same efficiency = $\frac{1}{120}$

Work done on

Day 1 = $1\times \frac{1}{120}$

Day 2 = $2\times \frac{1}{120}$

...

Day n = $\mathrm{n}\times \frac{1}{120}$

∴ Total Work = $1\times \frac{1}{120}$ + $2\times \frac{1}{120}$ + ... + $\mathrm{n}\times \frac{1}{120}$

This is also equal to 1.

∴ $\frac{1}{120}+\frac{2}{120}+\frac{3}{120}+..+\frac{n}{120}$ = 1

∑n = 120

⇒ $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ = 120

⇒ n = 15

Hence, 15.

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**3. CAT 2017 QA Slot 1 | Arithmetic - Average**

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

Answer: 11

**Explanation** :

There must be at least 1 person with weight 57 kg.

Remaining persons weigh a total of 630 - 57 = 573.

Maximum number of people in the group can be 573/53 + 1 (one who's weight is 57 kg) i.e. 10 + 1 = 11.

Hence, 11.

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**4. CAT 2017 QA Slot 1 | Arithmetic - Time, Speed & Distance**

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is:

Answer: 20

**Explanation** :

The speed in the second case is 5/4 times the speed in the first case.

Therefore, the time would be 4/5 times the time, i.e., 1/5 less.

This one fifth is 20 min. Therefore, the time taken in the first case is 100 min i.e. 100/60 = 5/3 hours.

The distance = 12 × 5/3 = 20 km.

Hence, 20.

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**5. CAT 2017 QA Slot 1 | Arithmetic - Percentage**

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59,500, then Ravi's total monthly savings (in Rs) is:

Answer: 70000

**Explanation** :

Let the total monthly savings be S.

Investment in FD = 50% of S = 0.5S

Remaining savings = S - 0.5S = 0.5S

Investment in stocks = 30% of 0.5S = 0.15S

Total invested amount (FD + stocks) = 0.5S + 0.15S = 0.65S

Remaing amount that is invested in savings account = 0.35S.

⇒ 0.35S + 0.5S = 59,500

⇒ 0.85S = 59500.

⇒ S = 70,000.

Hence, 70,000.

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**6. CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?

- A.
Give a discount of 5% on retail price

- B.
Give a discount of 2% on retail price

- C.
Increase the retail price by 2%

- D.
Sell at retail price

Answer: Option D

**Explanation** :

Let the retail price (MRP) be 100.

Discount = 15

Selling price = 85

Cost price = 85/1.02 = 500/6

In order to make a profit of 20%, the selling price = 500/6 × 1.2 = 100

The seller must sell at the retail price.

Hence, option (d).

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**7. CAT 2017 QA Slot 1 | Arithmetic - Time, Speed & Distance**

A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:

- A.
√6 : √2

- B.
√7 : 2

- C.
2√5 : 3

- D.
3 : 2

Answer: Option B

**Explanation** :

Let the speed of the boat in still water and the speed of the river be x and y respectively.

And the distance between office and home be 'd'.

⇒ $\frac{d}{2x+y}$ + $\frac{d}{2x-y}$ = $\frac{1}{4}\left(\frac{d}{x+y}+\frac{d}{x-y}\right)$

⇒ $\frac{d\left(4x\right)}{4{x}^{2}-{y}^{2}}$ = $\frac{1}{4}\left(\frac{d\left(2x\right)}{{x}^{2}-{y}^{2}}\right)$

⇒ 8(x^{2} – y^{2}) = 4x^{2} – y^{2}

⇒ x^{2}/y^{2} = 7/4

⇒ x/y = √7/2

Hence, option (b).

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**8. CAT 2017 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

Suppose, C_{1}, C_{2}, C_{3}, C_{4}, and C_{5} are five companies. The profits made by C_{1}, C_{2}, and C_{3} are in the ratio 9 : 10 : 8 while the profits made by C_{2}, C_{4}, and C_{5} are in the ratio 18 : 19 : 20. If C_{5} has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is:

- A.
438 crore

- B.
435 crore

- C.
348 crore

- D.
345 crore

Answer: Option A

**Explanation** :

C_{5} – C_{1} = 19.

∴ 100x - 81x = 19

⇒ x = 1

∴ Total profit = 81x + 90x + 72x + 95x + 100x = 438x = 438 crore.

Hence, option (a).

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**9. CAT 2017 QA Slot 1 | Arithmetic - Percentage**

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is:

- A.
35

- B.
50

- C.
60

- D.
65

Answer: Option D

**Explanation** :

Let the number of boys appearing for the admission test be b.

Number of girs appearing for the admission test = 2b.

Total boys who got admission = 45% of b = 0.45b

Total girls who got admission = 30% of 2b = 0.6b

Total students who got admission = 0.45b + 0.6b = 1.05b

∴ Percentage of candidates who get admission = $\frac{1.05b}{2b+b}\times 100\%$ = 35%

∴ 65% of the candidates do not get admission.

Hence, option (d).

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**10. CAT 2017 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio:

- A.
1 : 1

- B.
8 : 7

- C.
4 : 3

- D.
6 : 5

Answer: Option A

**Explanation** :

Total number of popcorn packets in stock = total number of chips packets = T

**Popcorn**:

Large : Super : Jumbo = 7 : 17 : 16

⇒ Jumbo popcorn packets = $\frac{16}{40}\mathrm{T}$

**Chips**:

Large : Super : Jumbo = 6 : 15 : 14

⇒ Jumbo popcorn packets = $\frac{14}{35}\mathrm{T}$

∴ Required ratio = $\frac{16}{40}\mathrm{T}$ : $\frac{14}{35}\mathrm{T}$ = 1 : 1.

Hence, option (a).

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**11. CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is:

- A.
6%

- B.
8%

- C.
10%

- D.
12%

Answer: Option B

**Explanation** :

Let the price of each good mango be g.

Price of each medium quality mango = g/2

Total cost price = 80g + 40(g/2) = 100g

Total selling price = 120(0.9g) = 108g

Overall profit = 8%

hence, option 2.

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**12. CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

- A.
30%

- B.
25%

- C.
24%

- D.
28%

Answer: Option D

**Explanation** :

Let the printed price be p.

If 40% discount is given, selling price = 0.6 × 60p = 36 p

20% profit is then made.

Total cost price = 36p/1.2 = 30p.

Ten toys are destroyed in the fire.

The remaining toys are sold at a price such that the same amount of profit is made as in the conditional case.

Profit made on remaining toys = 6p

Total selling price of remaining toys = 36p

Discount that should be given = 50p – 36p = 14p

Discount% = $\frac{14p}{50p}\times 100$% = 28%

Hence, option (d).

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**13. CAT 2017 QA Slot 1 | Algebra - Surds & Indices**

If a and b are integers of opposite signs such that (a + 3)^{2} : b^{2} = 9 : 1 and (a - 1)^{2}: (b - 1)^{2} = 4 : 1, then the ratio a^{2} : b^{2} is:

- A.
9 : 4

- B.
81 : 4

- C.
1 : 4

- D.
25 : 4

Answer: Option D

**Explanation** :

${\left(\frac{a+3}{b}\right)}^{2}=9$ and ${\left(\frac{a-1}{b-1}\right)}^{2}=4$

⇒ $\frac{\mathrm{a}+3}{\mathrm{b}}=\pm 3$ and $\frac{\mathrm{a}-1}{\mathrm{b}-1}=\pm 2$

We get following cases:

**Case 1(a)**: a + 3 = 3b and a - 1 = 2b - 2

⇒ a = 3 and b = 2 (Rejected)

**Case 1(b)**: a + 3 = 3b and a - 1 = -2b + 2

⇒ b is not an integer (Rejected)

**Case 2(a)**: a + 3 = -3b and a - 1 = 2b - 2

⇒ b is not an integer (Rejected)

**Case 2(b)**: a + 3 = -3b and a - 1 = -2b + 2

⇒ a = 15 and b = -6

∴ (a/b)^{2} = (15/-6)^{2} = 25/4

Hence, option (d).

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**14. CAT 2017 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is:

- A.
9.5

- B.
10

- C.
4.5

- D.
6

Answer: Option A

**Explanation** :

Let the average score of the boys in the midsemester examination be b.

Average score of the girls = b + 5

Average of the class = $\frac{20\mathrm{b}+30(\mathrm{b}+5)}{50}$ = b + 3

New average of girls = (b + 5) - 3 = b + 2

Let new average of the boys = x

Average score of the entire class increased by 2 and is hence (b + 3) + 2 = b + 5

∴ $\frac{20\mathrm{x}+30(\mathrm{b}+2)}{50}$ = b + 5

⇒ 20x + 30b + 60 = 50b + 250

⇒ 20x = 20b + 190

⇒ x = b + 9.5

Increase in the average of boys is 9.5

Hence, option (a).

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**15. CAT 2017 QA Slot 1 | Geometry - Coordinate Geometry**

The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is

- A.
4π

- B.
4

- C.
8

- D.
2π

Answer: Option C

**Explanation** :

The closed region bounded by |ax| + |by| = c in the two-dimensional plane has x-intercepts of ± $\frac{\mathrm{c}}{\left|\mathrm{a}\right|}$ and y-intercepts of ± $\frac{\mathrm{c}}{\left|\mathrm{b}\right|}$. This is in general a rhombus. In the given question, we have a square which has each of its diagonals as 4.

Area = 1/2 × 4 × 4 = 8.

Hence, option (c).

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**16. CAT 2017 QA Slot 1 | Geometry - Triangles**

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is:

- A.
225√3

- B.
500/√3

- C.
275√3

- D.
250√3

Answer: Option B

**Explanation** :

The medians of a triangle divide the triangle into six parts of equal area.

Area of the triangle) = $\sqrt{s(s-a)(s-b)(s-c)}$ = $250\sqrt{3}$

Area of GBC = $\frac{1}{3}$(Area of the triangle)

∴ Area of the remaining portion = 2/3 × 250√3 = 500/√3

Hence, option (b).

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**17. CAT 2017 QA Slot 1 | Geometry - Triangles | Geometry - Circles**

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq. cm, of the region enclosed by BPC and BQC is:

- A.
9π - 18

- B.
18

- C.
9π

- D.
9

Answer: Option B

**Explanation** :

AB = a (a = 6)

BC = a√2

BCQB is a semicircle of radius = half of BC = $\frac{\mathrm{a}}{\sqrt{2}}$

Area of semicircle BCQB = ½ × π × ${\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^{2}$ = ¼ × π × a^{2}

ACPBA is a quarter circle (quadrant) of radius a.

Area of BCPB = Area of sector ABPCA - Area of triangle ABC

= ¼ × π × a^{2} - Area of triangle ABC

Now, Area of CPBQC = Area of semicircle CQB - Area of CPBC

= ¼ × π × a^{2} - ( ¼ × π × a^{2} - Area of triangle ABC)

= Area of triangle ABC

∴ Area of region enclosed by CPBQC = Area of ∆ABC = ½ × 6 × 6 = 18.

Hence, option (b).

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**18. CAT 2017 QA Slot 1 | Geometry - Mensuration**

A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to:

- A.
10

- B.
50

- C.
60

- D.
20

Answer: Option B

**Explanation** :

The volumes of the 5 smaller cubes and the original big one are in the ratio 1 : 1 : 8 : 27 : 27 : 64.

Therefore, the sides are in the ratio 1 : 1 : 2 : 3 : 3 : 4 while the areas are in the ratio 1 : 1 : 4 : 9 : 9 : 16.

The sum of the areas of the 5 smaller cubes is 24 parts while that of the big cube is 16 parts.

The sum is 50% greater.

Hence, option (b).

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**19. CAT 2017 QA Slot 1 | Geometry - Mensuration**

A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9π cm^{3}. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is:

Answer: 6

**Explanation** :

The height of the cylinder (h) = 3

The volume = 9π

⇒ πr^{2}h = 9π

⇒ r = √3

The radius of the ball (R) = 2

The height of O, the centre of the ball, above the line representing the top of the cylinder is say a.

Hence, 2^{2} = a^{2} + (√3)^{2}

∴ a = 1

∴ The height of the topmost point of the ball from the base of the cylinder is = height of cylinder + height of center of ball from the top of cylinder + radius of the ball = 3 + 1 + 2 = 6.

Hence, 6.

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**20. CAT 2017 QA Slot 1 | Geometry - Triangles**

Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is:

Answer: 24

**Explanation** :

Given triangle is right triangle with smaller sides as 15 and 20, hence the hypotenuse will be 25.

Altitude from right vertex to hypotenuse = (15 × 20)/25 = 12

This is the shortest distance from A to BC. At 30 km/hr = 12/30 hrs = 12/30 × 60 mins = 24 mins.

Hence, 24.

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**21. CAT 2017 QA Slot 1 | Algebra - Progressions | Algebra - Logarithms**

Suppose, log_{3}x = log_{12}y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log_{6}G is equal to:

- A.
√a

- B.
2a

- C.
a/2

- D.
a

Answer: Option D

**Explanation** :

log_{3}x = a ⇒ x = 3^{a}.

log_{12}y = a ⇒ y = 12^{a}.

∴ xy = 36^{a} and √(xy) = G = 6^{a}.

∴ log_{6}G = a.

Hence, option (d).

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**22. CAT 2017 QA Slot 1 | Algebra - Number Theory**

If x + 1 = x^{2} and x > 0, then 2x^{4} is:

- A.
6 + 4√5

- B.
3 + 5√5

- C.
5 + 3√5

- D.
7 + 3√5

Answer: Option D

**Explanation** :

x + 1 = x^{2}

⇒ x^{2 }- x – 1 = 0

⇒ x = (1+√5)/2 (∵ x > 0)

Now, squaring the original equation, we get

(x +1)^{2} = x^{4}

⇒ x^{2} + 1 + 2x = x^{4}

⇒ x^{4} = x + 1 + 1 + 2x (Since, x^{2} = x + 1)

⇒ x^{4} = 3x + 2

⇒ 2x^{4} = 6x + 4

⇒ 2x^{4} = 6((1+√5)/2) + 4

⇒ 2x^{4} = 7 + 3√5

**Alternately**,

We know, x = (1+√5)/2

⇒ x ≈ 1.62

∴ 2x^{4} ≈ 13.8

option I ≈ 14.8,

option II ≈ 11.6,

option IV ≈13.6

Hence, option (d).

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**23. CAT 2017 QA Slot 1 | Algebra - Logarithms**

The value of log_{0.008}√5 + log_{√3}81 – 7 is equal to:

- A.
1/3

- B.
2/3

- C.
5/6

- D.
7/6

Answer: Option C

**Explanation** :

0.008 = 8/1000 = 1/125 = 5^{-3}

∴ log_{0.008}√5 = ${\mathrm{log}}_{{5}^{-3}}\left({5}^{1/2}\right)$ = $\frac{1/2}{-3}$ = $\frac{-1}{6}$ and

log_{√3}81 = ${\mathrm{log}}_{{3}^{1/2}}\left({3}^{4}\right)$ = $\frac{4}{1/2}$ = 8.

∴ The given expression is -1/6 + 8 - 7 = 5/6

Hence, option (c).

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**24. CAT 2017 QA Slot 1 | Algebra - Surds & Indices**

If 9^{2x–1} – 81^{x-1} = 1944, then x is

- A.
3

- B.
9/4

- C.
4/9

- D.
1/3

Answer: Option B

**Explanation** :

9^{2x–1} – 81^{x-1} = 1944

⇒ 9^{2x–2 }× 9 – 9^{2x-2} = 1944

⇒ 9^{2x–2} (9 – 1) = 8 × 243 = 8 × 3^{5}

⇒ 3^{4x–4} × 8 = 8 × 3^{5}

⇒ ∴ 4x – 4 = 5

∴ x = 9/4.

Hence, option (b).

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**25. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

- A.
101

- B.
99

- C.
87

- D.
105

Answer: Option B

**Explanation** :

x = 25 + y + z. The possible values of x, y, z and the corresponding number of values of y, z are tabulated below (x, y, z are positive integers).

We see that 27 ≤ x ≤ 40.

The number of solutions is 1 + 2 + … + 12 + 11 + 10 = 78 + 21 = 99.

Hence, option (b).

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**26. CAT 2017 QA Slot 1 | Algebra - Inequalities & Modulus**

For how many integers n, will the inequality (n – 5) (n – 10) – 3(n – 2) ≤ 0 be satisfied?

Answer: 11

**Explanation** :

(n – 5) (n – 10) – 3(n – 2) ≤ 0

⇒ n^{2} – 15n + 50 - 3n + 6 ≤ 0

⇒ n^{2} – 18n + 56 ≤ 0

⇒ (n – 4) (n – 14) ≤ 0

As n is an integer, n can be 4, 5, 6 ……14, i.e. it can have 11 values.

Hence, 11.

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**27. CAT 2017 QA Slot 1 | Algebra - Quadratic Equations**

If f_{1}(x) = x^{2} + 11x + n and f_{2}(x) = x, then the largest positive integer n for which the equation f_{1}(x) = f_{2}(x) has two distinct real roots, is :

Answer: 24

**Explanation** :

x^{2} + 11x + n = x

⇒ x^{2} + 10x + n = 0

For roots to be real and distinct, Discriminant > 0

⇒ D = (10)^{2} - 4 × 1 × n > 0

⇒ 4n < 100

⇒ n < 25

Therefore, highest possible integral value of n is 24.

Hence, 24.

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**28. CAT 2017 QA Slot 1 | Algebra - Inequalities & Modulus**

If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b)^{2} + (a - c)^{2} + (a - d)^{2} is

Answer: 2

**Explanation** :

a + b + c + d = 30, a, b, c, d are integers.

(a – b)^{2} + (a – c)^{2} + (a – d)^{2}would have its maximum value when each bracket has the least possible value.

Let (a, b, c, d) = (8, 8, 7, 7)

The given expression would be 2. It cannot have a smaller value.

Hence, 2.

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**29. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Answer: 160

**Explanation** :

There are 5 pairs of diametrically opposite points and the centre O.

If O is not selected, the number of triangles = ^{10}C_{3} = 120.

If O is selected, the other two points can be selected in ^{10}C_{2} - 5 = 40 ways. (when 3 points on a diameter are selected we will not get a triangle)

The number of triangles is = 120 + 40 = 160.

**Alternately**,

No. of ways of choosing 3 points out of given 11 points = ^{11}C_{3} = 165.

Out of these 165 ways, 5 of these would give us 3 points on the diameters mentioned, which will not form a triangle.

Hence, no. of triangles = 165 - 5 = 160.

Hence, 160.

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**30. CAT 2017 QA Slot 1 | Algebra - Functions & Graphs**

The shortest distance of the point (1/2,1) from the curve y = |x - 1| + |x + 1| is

- A.
1

- B.
0

- C.
√2

- D.
√32

Answer: Option A

**Explanation** :

The graph of y = |x – 1| + |x + 1| is shown above.

The shortest distance of (1/2, 1) from the graph is 1.

Hence, option (a).

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**31. CAT 2017 QA Slot 1 | Algebra - Progressions**

If the square of the 7^{th} term of an arithmetic progression with positive common difference equals the product of the 3^{rd} and 17^{th} terms, then the ratio of the first term to the common difference is:

- A.
2 : 3

- B.
3 : 2

- C.
3 : 4

- D.
4 : 3

Answer: Option A

**Explanation** :

Let the first term be a and the common difference be d.

(a + 6d)^{2} = (a + 2d) (a + 16d)

⇒ a^{2} + 12ad + 36d^{2} = a^{2} + 18ad + 32d^{2}

⇒ 4d^{2} = 6ad

⇒ a/d = 2/3

Hence, option (a).

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**32. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

- A.
16

- B.
20

- C.
14

- D.
15

Answer: Option A

**Explanation** :

After giving one eraser to each of the 4 kids, there are 3 left.

They can split 2, 1 or 1, 1, 1. (No kid can get 4)

Number of ways of 2, 1 split = ^{4}P_{2}

Number of ways of 1, 1, 1 split = ^{4}C_{3}

There are ^{4}P_{2} + ^{4}C_{3}, i.e., 16 ways of distributing the erasers.

**Alternately,**

Let the number of erasers given to the 4 kids be w, x, y, z.

w + x + y + z = 7.

After giving at least 1 eraser to each, 3 erasers will be left.

w' + x' + y' + z' = 3

The number of positive integral solutions is ^{3 + 4 - 1}C_{4 - 1}, i.e. 20. This includes (4,1,1,1); (1,4,1,1); (1,1,4,1); (1,1,1,4).

The required number = 20 – 4 = 16.

Hence, option (a).

Workspace:

**33. CAT 2017 QA Slot 1 | Algebra - Functions & Graphs**

If f(x) = (5x+2)/(3x-5) and g(x) = x^{2} – 2x – 1, then the value of g(f(f(3))) is:

- A.
2

- B.
13

- C.
6

- D.
23

Answer: Option A

**Explanation** :

f(x) = (5x+2)/(3x-5), g(x) = x^{2} – 2x – 1

f(3) = $\frac{5\times 3+2}{3\times 3-5}=\frac{17}{4}$

f(17/4) = $\frac{5\times {\displaystyle \frac{17}{4}}+2}{3\times {\displaystyle \frac{17}{4}}-5}$ = $\frac{85+8}{51-20}$ = $\frac{93}{31}$ = 3

g(3) = 3^{2} – 2 × 3 – 1 = 2.

Hence, option (a).

Workspace:

**34. CAT 2017 QA Slot 1 | Algebra - Progressions**

Let a_{1}, a_{2},.......a_{3n} be an arithmetic progression with a_{1} = 3 and a_{2} = 7. If a_{1} + a_{2} + ......+ a_{3n} = 1830, then what is the smallest positive integer m such that m(a_{1} + a_{2} + ..... + a_{n}) > 1830?

- A.
8

- B.
9

- C.
10

- D.
11

Answer: Option B

**Explanation** :

a_{1} = 3, a_{2} = 7, …..

a_{n} = 3 + (n - 1) × 4 = 4n – 1,

a_{3n} = 3 + (3n – 1) × 4 = 12n - 1

a_{1} + a_{2} + a_{3} + … + a_{3n} = $\frac{3\mathrm{n}}{2}(12\mathrm{n}-1+3)$ = 1830

⇒ n(6n + 1) = 610

⇒ 6n + n – 610 = 0

⇒ (6n + 61)(n - 10) = 0

⇒ n - 10 = 0

Now a_{10} = 3 + (10 - 1) × 4 = 39

∴ a_{1} + a_{2} + a_{3} + … + a_{10} = 3 + 7 + … + 39 = $\frac{10}{2}(3+39)$ = 210.

210 × m > 1830

⇒ n > 1830/210 = 8.7.

The minimum integral value of m is 9.

Hence, option (b).

Workspace:

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